A) \[\frac{1}{4}{{t}_{m}}\] is the arithmetic mean of \[{{t}_{m-1}}\] and \[{{t}_{m-2}}\]
B) \[\frac{1}{4}{{t}_{m-1}}\] is the arithmetic mean of \[{{t}_{m}}\] and \[{{t}_{m-2}}\]
C) \[\frac{1}{4}{{t}_{m}}\] is the geometric mean of \[{{t}_{m-1}}\] and \[{{t}_{m-2}}\]
D) \[\frac{1}{4}{{t}_{m-1}}\] is the geometric mean of \[{{t}_{m}}\] and \[{{t}_{m-2}}\]
Correct Answer: A
Solution :
Given, \[{{t}_{n}}=\frac{1}{2}\{{{(1+\sqrt{3})}^{n}}+{{(1-\sqrt{3})}^{n}}\},\,n=3,4,5,....\] Now, \[{{t}_{m}}=\frac{1}{2}\{{{(1+\sqrt{3})}^{m}}+{{(1-\sqrt{3})}^{m}}\}\] AM of \[{{t}_{m-1}}\] and \[{{t}_{m-2}}\] is \[\frac{{{t}_{m-1}}\,+{{t}_{m-2}}}{2}\] \[\therefore \] \[\frac{\begin{align} & \frac{1}{2}\{{{(1+\sqrt{3})}^{m-1}}+{{(1-\sqrt{3})}^{m-1}}+{{(1+\sqrt{3})}^{m-2}} \\ & +{{(1-\sqrt{3})}^{m-2}} \\ \end{align}}{2}\] \[=\frac{1}{4}\{{{(1+\sqrt{3})}^{m-1}}+{{(1+\sqrt{3})}^{m-2}}+{{(1-\sqrt{3})}^{m-1}}\] \[+{{(1-\sqrt{3})}^{m-2}}\}\] \[=\frac{1}{4}\{{{(1+\sqrt{3})}^{m-2}}(1+\sqrt{3}+1)+{{(1-\sqrt{3})}^{m-2}}(1-\sqrt{3}+1)\}\]\[=\frac{1}{4}\{{{(1+\sqrt{3})}^{m-2}}(2+\sqrt{3})+{{(1-\sqrt{3})}^{m-2}}(2-\sqrt{3})\}\]\[=\frac{1}{4}\left[ \frac{1}{2}\{{{(1+\sqrt{3})}^{m-2}}(4+2\sqrt{3})+{{(1-\sqrt{3})}^{m-2}}(4-2\sqrt{3}\} \right]\] \[=\frac{1}{4}\left[ \frac{1}{2}\{{{(1+\sqrt{3})}^{m-2}}{{(1+\sqrt{3})}^{2}}+{{(1-\sqrt{3})}^{m-2}}{{(1-\sqrt{3})}^{2}}\} \right]\]\[=\frac{1}{4}\left[ \frac{1}{2}\{{{(1+\sqrt{3})}^{m}}+{{(1-\sqrt{3})}^{m}}\} \right]=\frac{1}{4}{{t}_{m}}\]
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